Question

A particle is projected at ${60}^{}$ to the horizontal with a kinetic energy KE. The kinetic energy at the highest point is :

• A. KE
• B. Zero
• C. $\frac{KE}{4}$
• D. $\frac{KE}{2}$

Answer 1

The correct option is C $\frac{KE}{4}$

Let initial velocity of the particle be $v$.
$\therefore KE=\frac{1}{2}{mv}^2$
at the highest point of its flight
$v=u\cos {60}^{}=\frac{u}{2}$
$\therefore$ Kinetic energy at this point
$=\frac{1}{2}{mv}^2=\frac{1}{2}\frac{{mu}^2}{4}=\frac{KE}{4}$